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Math help...(Epsilon-Delta definition of a limit)

eLiu

Diamond Member
Jun 4, 2001
6,407
1
0
Hey guys...I know there are some math-majors on here, so hopefully someone can give me a push in the right direction. Note: this is not for homework--I plan on taking this class (Complex Analysis) in the fall, and I'm trying to learn some of the material ahead of time to make it easier.

So...the problem... (E denotes epsilon, D delta)
For anyone that may need a reminder...what we're trying to show is the limit of f(z) = w0 as z->z0 by demonstrating that for any E>0 there exists a positive D such that:
|f(z) - w0| < E whenver 0 < |z-z0| < D.

The limit i'm trying to prove: limit of 1/z = i as z->-i.
So that gives me: |1/z - i| < E whenever |z+i|<D.

Then I know I need to try and get the |1/z-i| in terms of |z+i|...but that's where my hang-up is.

Through some manipulations...I've gotten that |1/z-i| = |z+i|/|z|...but there's still that annoying z term.

If I change it to |z+i|/|z+i-i|, I can do the triangle inequality in reverse and get:

|z+i|/|z+i-i| <= |z+i|/(|z+i|-1).

And there...I'm stuck. Trying to say D=E/k and substitute that into the above doesn't work out too good. Dividing top/bottom by |z+i| gets me something in the form of 1/1-x...but the power-expansion doesn't converge >.<

So with that, I've just about run out of ideas... Do I even have the right approach?? I don't want the whole proof...just a hint.

Thanks,
-Eric
 

Kyteland

Diamond Member
Dec 30, 2002
5,747
1
0
I'll give you a bump because I don't want to think about it right now. :)

Maybe I'll answer when I get home.
 

TuxDave

Lifer
Oct 8, 2002
10,577
3
71
:confused:

I haven't done this for a while. Can you give a very basic proof using delta epsilon? Like proving something stupid like as x approaches 0, x^2 goes to 0. I just need to be reminded of the basic idea.
 

eLiu

Diamond Member
Jun 4, 2001
6,407
1
0
Originally posted by: TuxDave
:confused:

I haven't done this for a while. Can you give a very basic proof using delta epsilon? Like proving something stupid like as x approaches 0, x^2 goes to 0. I just need to be reminded of the basic idea.
Well hm, here's a proof I just did...see if it helps:
limit as z->i of z^2 = -1.

so, establish initial conditions: |z^2 + 1| < E whenever |z-i| < D

Keeping in mind that we want to try and find |z^2+1| in terms of |z-i|:

|z^2+1| = |z-i||z+i|=|z-i||z-i+2i|
Then, by the triangle inequality: |a+b|<=|a|+|b|:
|z-i||z-i+2i|< |z-i|(|z-i|+|2i|) = |z-i|(|z-i| + 2)

Next...remember the objetive is to relate D to E...so say D=E/k, and now we need to find a value for E.

Ok now, multiply that out, you get |z-i|^2 + 2|z-i|. Remembering that |z-i| is small, that means the squared term is insignificant b/c it vanishes compared to the linear term. Thus, going back to |z-i|(|z-i| + 2), we can safely substitute anything for the |z-i| term inside the ()--b/c it vanishes.

For the outside |z-i|, we substitute in e/k. For the inside one, we substitute in 1 (to try--we could start with 10,000, but we want the smallest value, and the convention is to use the natural nubmers...b/c like 0.057285 would be annoying).

That results in this: |z-i|(|z-i| + 2) < (E/k)(1+2) (It's a less-than relation b/c |z-i| is less than D..so when you substitute E/k in, you get a number that's bigger than what you started with)

Choosing k=3, we have: |z-i|(|z-i| + 2) < 3*E/3 = E
Thus:
|z^2+1| < |z-i|(|z-i| + 2) < E.

And that does it :)
 

EngenZerO

Diamond Member
Dec 24, 2001
5,099
1
0
dont you have all summer to learn? let me guess your the type of person who likes to destroy curves, ;).

I have a proof in my math notes but I am too lazy to go find them in my basement and it was three years ago when I took that class so I am a tad hazy, good luck.
 

eLiu

Diamond Member
Jun 4, 2001
6,407
1
0
Originally posted by: EngenZerO
dont you have all summer to learn? let me guess your the type of person who likes to destroy curves, ;).

I have a proof in my math notes but I am too lazy to go find them in my basement and it was three years ago when I took that class so I am a tad hazy, good luck.
haha, yeah i do...but still, I need something to do now instead of spending all my time gaming :)

Though I think I've found a solution using the mclaurin expansion of 1/(1+x)...I'm gonna check it with a teacher at my HS who still remembers this stuff to see if it's valid though...

-Eric

Edit: And sadly, where I'm going...I probably won't be setting any curves--I'll be complaining about other people doing it...
 

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